ISI MStat PSB 2009 Problem 1 | Nilpotent Matrices

This is a very simple sample problem from ISI MStat PSB 2009 Problem 1. It is based on basic properties of Nilpotent Matrices and Skew-symmetric Matrices. Try it !

Problem- ISI MStat PSB 2009 Problem 1


(a) Let \(A\) be an \(n \times n\) matrix such that \((I+A)^4=O\) where \(I\) denotes the identity matrix. Show that \(A\) is non-singular.

(b) Give an example of a non-zero \(2 \times 2\) real matrix \(A\) such that \( \vec{x'}A \vec{x}=0\) for all real vectors \(\vec{x}\).

Prerequisites


Nilpotent Matrix

Eigenvalues

Skew-symmetric Matrix

Solution :

The first part of the problem is quite easy,

It is given that for a \(n \times n\) matrix \(A\), we have \((I+A)^4=O\), so, \(I+A\) is a nilpotet matrix, right !

And we know that all the eigenvalues of a nilpotent matrix are \(0\). Hence all the eigenvalues of \(I+A\) are 0.

Now let \(\lambda_1, \lambda_2,......,\lambda_k\) be the eigenvalues of the matrix \(A\). So, the eigenvalues of the nilpotent matrix \(I+A\) are of form \(1+\lambda_k\) where, \(k=1,2.....,n\). Now since, \(1+\lambda_k=0\) which implies \(\lambda_k=-1\), for \(k=1,2,...,n\).

Since all the eigenvalues of \(A\) are non-zero, infact \(|A|=(-1)^n \). Hence our required propositon.

(b) Now this one is quite interesting,

If for any \(2\times 2\) matrix, the Quadratic form of that matrix with respect to a vector \(\vec{x}=(x_1,x_2)^T\) is of form,

\(a{x_1}^2+ bx_1x_2+cx_2x_1+d{x_2}^2\) where \(a,b,c\) and \(d\) are the elements of the matrix. Now if we equate that with \(0\), what condition should it impose on \(a, b, c\) and \(d\) !! I leave it as an exercise for you to complete it. Also Try to generalize it you will end up with a nice result.


Food For Thought

Now, extending the first part of the question, \(A\) is invertible right !! So, can you prove that we can always get two vectors from \(\mathbb{R}^n\), say \(\vec{x}\) and \(\vec{y}\), such that the necessary and sufficient condition for the invertiblity of the matrix \(A+\vec{x}\vec{y'}\) is " \(\vec{y'} A^{-1} \vec{x}\) must be different from \(1\)" !!

This is a very important result for Statistics Students !! Keep thinking !!


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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TIFR 2017 Math Solution -Set of Nilpotent Matrices


TIFR 2017 Math Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Linear Algebra by Gilbert Strang. This book is very useful for the preparation of TIFR Entrance.

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Problem:State True or False


The set of nilpotent matrices of ( M_3 (\mathbb{R} ) ) spans ( M_3 (\mathbb{R} ) ) considered as an ( \mathbb {R} ) - vector space ( a matrix A is said to be nilpotent if there exists ( n \in \mathbb{N} )  such that ( A^n = 0 ) ).


Discussion


If such a basis exist. Then it will contain 9 matrices (because dimension of ( M_3 ( \mathbb{R} ) ) is 9 ).

Suppose ( { N_1, ... , N_9 } ) be the 9 nilpotent matrices which span ( M_3 (\mathbb{R} ) ). Now consider the usual basis of ( M_3 (\mathbb{R} ) ) :

$$ M_1 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} , M_2 = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix} , ... , M_9 = \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}$$

There is a linear transformation L (via a ( 9 \times 9 )  change of basis matrix) that sends ( { N_1, ... , N_9 } ) to ( { M_1, ... , M_9 } ).


Chatuspathi

TIFR 2014 Problem 11 Solution - Nilpotent Matrix Eigenvalues


TIFR 2014 Problem 11 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Introduction to Linear Algebra by Gilbert Strang. This book is very useful for the preparation of TIFR Entrance.

Also Visit: College Mathematics Program of Cheenta


Problem:


Let (A) be an (nxn) matrix with real entries such that (A^k=0) (0-matrix) for some (k\in\mathbb{N}).

Then

A. A has to be the 0-matrix.

B. trace(A) could be non-zero.

C. A is diagonalizable.

D. 0 is the only eigenvalue of A


Discussion:


Let (v) be an eigenvector of (A) with eigenvalue (\lambda).

Then (v \neq 0) and (Av=\lambda v).

Again, (A^2 v=A(Av)=A(\lambda v)=\lambda Av= (\lambda)^2v).

We continue to apply A, applying it k times gives: (A^k v=(\lambda)^k v).

By given information, the left hand side of the above equality is 0.

So (\lambda^k v=0) and remember (v \neq 0).

So (\lambda =0).

Therefore (0) is the only eigenvalue for (A).

So D is true.

We analyse the question a little bit further, to check it satisfies no other options above.

We know (trace(A)=) sum of eigenvalues of A= (\sum 0 =0)

So option B is false.

Take (A=\begin{bmatrix} 0 & 1 // 0 & 0 \end{bmatrix} ).

Then (A^2 =0). But (A) is not the zero matrix.

Also, if (A) were diagonalizable then the corresponding diagonal matrix would be the zero matrix. Which would then imply that (A) is the zero matrix, which in this case it is not. (See  TIFR 2013 Probmem 8 Solution-Diagonalizable Nilpotent Matrix ) So this disproves options A and C.


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